At the vertex, the tangent line to the graph will be horizontal, with slope 0. Therefore,
we search for an x such that f'(x) = 0. We have

f'(x)

=

limΔx→0

=

limΔx→0

=

limΔx→0

=

limΔx→02x + 2 + Δx

=

2(x + 1)

Thus f'(x) = 0 if and only if x = - 1. Now f (x) = (- 1)2 + 2(- 1) + 2 = 1, so the vertex of
the parabola is (- 1, 1). We can check this by noting that f (x) - 1 = (x + 1)2, so the graph
of f (x) is the graph of x2 translated 1 unit to the left and 1 unit up.

Problem :
Find the equation of the tangent line to the graph of f (x) = x3 at x = 2.

First we compute f'(2):

f'(2)

=

limΔx→0

=

limΔx→0

=

12

The equation of the line through (2, f (2)) = (2, 8) with slope 12 is given by y = 12(x - 2) + 8.